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I have a question about the irrationality of $e$:

In proving the irrationality of $e$, one can prove the irrationality of $e^{-1}$ by using the series $$e^x = 1+x+\frac{x^2}{2!} + \cdots + \frac{x^n}{n!}+ \cdots$$ So the series for $e^{-1}$ is $$s_n = \sum_{k=0}^{n} \frac{(-1)^{k}}{k!}$$ so that $$0 < e^{-1}-s_{2k-1} < \frac{1}{(2k)!}$$ or $$0 < (2k-1)!(e^{-1}-s_{2k-1}) < \frac{1}{2k} \leq \frac{1}{2}$$ If $e^{-1}$ were rational then we would have a difference of two integers which is not an integer (i.e. between $0$ and $\frac{1}{2}$).

Question: Is this "irrationality of $e^{-1}$ by alternating series" proof what motivated the definition of irrationality measure? It seems that this was borne out of using alternating series.

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I always thought the definition of irrationality measure was motivated from Liouville's theorem:… – Qiaochu Yuan Aug 3 '11 at 5:14
I think there is no connection. Incidentally, I prefer going through $e^{-1}$, as you did. Proofs I have seen of the irrationality of $e$ use the series for $e$ directly, which involves some additional work to produce the required estimate. – André Nicolas Aug 3 '11 at 5:30
This is a really nice proof that I hadn't seen before. Thanks! – ShreevatsaR Aug 3 '11 at 11:49
As far as I know, the "series proof" for the irrationality of $e$ first appeared in the following 1815 book by Janot de Stainville (Article 232, pp. 339-341): Stainville says he learned the proof from Poinsot and that the proof itself is due to Fourier. – Dave L. Renfro Aug 3 '11 at 14:16
J.H.Lambert (1728-1777) gave a proof (c.1766) that $\pi$ is irrational. In the early to mid19th century preceding the Hermite-Lindemann transcendence theorem there were limited results,e.g. : If $A,B,C$ are rational and $Ae^2+Be+C=0$ then $A=B=C=0$ (easy). If $r\in Q$ and $r\ne 0$ then $e^r\in Q$ (not easy). – user254665 Nov 29 '15 at 5:33

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